적분 전체 문제

The integrand is similar to $\dfrac{1}{\sqrt{1-x^2}}$. In order to formulate a guess for the antiderivative, let's factor out $\sqrt{2}$ from the denominator:

$$\begin{aligned} \displaystyle\int \dfrac{1}{\sqrt{2-x^2}}\mathrm{d}x&= \displaystyle\int \dfrac{1}{\sqrt{2\left(1-\frac{x^2}{2}\right)}}\mathrm{d}x\\ &=\displaystyle\int \dfrac{1}{ \sqrt{2}\sqrt{1-\frac{x^2}{2}}}\mathrm{d}x\\ &= \int\frac{1}{\sqrt{2}}\cdot \dfrac{1}{ \sqrt{1-\left(\dfrac{x}{\sqrt2}\right)^2}}\mathrm{d}x\\ \\ \text{At this point, we might guess that our antiderivative is something like $F(x) = \arcsin\left(\dfrac{x}{\sqrt{2}}\right)$. To explore this possibility, we can differentiate, and see what we get.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin\left(\dfrac{x}{\sqrt{2}}\right)\right\}&=\frac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{1-\left(\dfrac{x}{\sqrt{2}}\right)^2}} \\ \text{This is exactly what we want! So,} \\ \int \dfrac{1}{\sqrt{2-x^2}}\mathrm{d}x&=\arcsin\left(\dfrac{x}{\sqrt2}\right)+C \end{aligned}$$

We know that $\int \sec^2 x\ \mathrm{d}x = \tan x +C$, and $\sec^2x = \tan^2 x + 1$, so

$$\begin{aligned} \int \tan^2 x\ \mathrm{d}x &= \int \sec^2x - 1 \ \mathrm{d}x\\ &=\int \sec^2 x \ \mathrm{d}x - \int 1 \ \mathrm{d}x\\ &=\tan x - x + C \end{aligned}$$

• Solution 1: This might not obviously look like the derivative of anything familiar, but it does look like half of a familiar trig identity: $2\sin x \cos x = \sin(2x)$.

$$\begin{aligned} \int 3 \sin x \cos x \ \mathrm{d}x&=\int \frac{3}{2}\cdot 2\sin x \cos x \ \mathrm{d}x\\ &=\int \frac{3}{2} \sin(2x)\ \mathrm{d}x \\ \text{So, we might guess that the antiderivative is something like $-\cos(2x)$. We only need to figure out the constants.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{-\cos(2x)\}&=2\sin(2x)\\ \text{So,}\qquad \frac{\mathrm{d}}{\mathrm{d}x}\left\{-\frac{3}{4}\cos(2x)\right\}&=\frac{3}{2}\sin(2x)\\ \text{Therefore,}\qquad \int 3\sin x \cos x\ \mathrm{d}x&=-\frac{3}{4}\cos(2x)+C \end{aligned}$$

• Solution 2: You might notice that the integrand looks like it came from the chain rule, since $\cos x$ is the derivative of $\sin x$. Using this observation, we can work out the antideriative:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sin^2 x\right\}&=2\sin x \cos x\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{3}{2}\sin^2 x\right\}&=3\sin x \cos x\\ \text{So,}\qquad \int 3\sin x \cos x \ \mathrm{d}x&=\frac{3}{2}\sin^2 x+C \end{aligned}$$

These two answers look different. Using the identity $\cos(2x)=1 - 2 \sin^2(x)$, we reconcile them:

$$\begin{aligned} -\frac{3}{4}\cos(2x)+C&= -\frac{3}{4}\left(1-2\sin^2x\right)+C\\ &=\frac{3}{2}\sin^2 x + \left(C-\frac{3}{4}\right) \end{aligned}$$

The $\frac{3}{4}$ here is not significant. Remember that $C$ is used to designate a constant that can take any value between $-\infty$ and $+\infty$. So $C-\frac{3}{4}$ is also just a constant that can take any value between $-\infty$ and $+\infty$. As the two answers we found differ by a constant, they are equivalent.

It's not immediately obvious which function has $\cos^2 x$ as its derivative, but we can make the situation a little clearer by using the identity $\cos^2 x = \dfrac{1+\cos(2x)}{2}$:

$$\begin{aligned} \int \cos^2 x \ \mathrm{d}x&=\int \frac{1}{2}\cdot \left(1+\cos(2x)\right)\ \mathrm{d}x\\ &=\int \frac{1}{2}\ \mathrm{d}x + \int \frac{1}{2}\cos(2x)\ \mathrm{d}x\\ &=\frac{1}{2}x + C+ \int \frac{1}{2}\cos(2x)\ \mathrm{d}x \\ \text{For the remaining integral, we might guess something like $F(x) = \sin(2x)$. Let's figure out the appropriate constant:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sin(2x) \right\}&=2\cos(2x)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{1}{4}\sin(2x) \right\}&=\frac{1}{2}\cos(2x)\\ \text{So,}\qquad \int \frac{1}{2}\cos(2x)\ \mathrm{d}x &=\frac{1}{4}\sin(2x)+C\\ \text{Therefore,}\qquad \int \cos^2 x \ \mathrm{d}x&=\frac{1}{2}x+\frac{1}{4}\sin(2x)+C \end{aligned}$$

By the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} F'(x)&=\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x \log(2+\sin t)\,\mathrm{d}t &&=\log(2+\sin x) \\ G'(y)&=\frac{\mathrm{d}}{\mathrm{d}y}\bigg[- \int_0^y \log(2+\sin t)\,\mathrm{d}t\bigg] &&=-\log(2+\sin y) \\ \end{aligned}$$

So,

$$F'\Big(\frac{\pi}{2}\Big)=\log 3\qquad G'\Big(\frac{\pi}{2}\Big)=-\log(3)$$

By the Fundamental Theorem of Calculus Part 1,

$$f'(x) = 100(x^2-3x+2)e^{-x^2} = 100(x-1)(x-2)e^{-x^2}$$

As $f(x)$ is increasing whenever $f'(x)>0$ and $100 e^{-x^2}$ is always strictly bigger than $0$, we have $f(x)$ increasing if and only if $(x-1)(x-2)>0$, which is the case if and only if $(x-1)$ and $(x-2)$ are of the same sign. Both are positive when $x>2$ and both are negative when $x<1$. So $f(x)$ is increasing when $-\infty<x<1$ and when $2<x<\infty$.

Remark: even without the Fundamental Theorem of Calculus, since $f(x)$ is the area under a curve from 1 to $x$, $f(x)$ is increasing when the curve is above the $x$-axis (because we're adding positive area), and it's decreasing when the curve is below the $x$-axis (because we're adding negative area).

Applying $\frac{\mathrm{d}}{\mathrm{d}x}$ to both sides of $x^2=1+\int_1^x f(t)\ \mathrm{d}t$ gives, by the Fundamental Theorem of Calculus Part 1, $2x=f(x)$.

Apply $\frac{\mathrm{d}}{\mathrm{d}x}$ to both sides of $x \sin(\pi x) = \int_0^x f(t)\, \mathrm{d}t$. Then, by the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} &&f(x) =\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f(t)\, \mathrm{d}t&=\frac{\mathrm{d}}{\mathrm{d}x}\big\{x \sin(\pi x)\big\} \\ \implies&& f(x) &= \frac{\mathrm{d}}{\mathrm{d}x}\big\{x \sin(\pi x)\big\} =\sin(\pi x)+\pi x\cos(\pi x) \\ \implies&& f(4)&= \sin(4\pi)+4\pi\cos(4\pi)=4\pi \end{aligned}$$

The car stops when $v(t)=30-10t=0$, which occurs at time $t = 3$. The distance covered up to that time is

$$\begin{aligned} \int_0^3 v(t)\,\mathrm{d}t = (30t - 5t^2)\Big|_{0}^{3} = (90-45)-0 = 45\,\textrm{m}. \end{aligned}$$

(a) Using the product rule, followed by the chain rule, followed by the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} f'(x) & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x^3+1} e^{t^3} \mathrm{d}t \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] \Bigg[\frac{d}{\mathrm{d}y}\int_{0}^{y} e^{t^3} \mathrm{d}t\Bigg]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] \Big[e^{y^3}\Big]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] e^{(x^3+1)^3} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + 3x^5 e^{(x^3+1)^3} \end{aligned}$$

(b) In general, the equation of the tangent line to the graph of $y=f(x)$ at $x=a$ is

$$y=f(a) + f'(a)\,(x-a)$$

Substituting in the given $f(x)$ and $a=-1$:

$$\begin{aligned} f(a)=f(-1)&=(-1)^3\int_0^0 e^{t^3}\ \mathrm{d}t=0\\ f'(a)=f'(-1)&=3(-1)^2\int_{0}^{0}e^{t^3}\ \mathrm{d}t + 3(-1)^5e^{0}\\ &=0-3=-3\\ (x-a)=x-(-1)&=x+1 \\ \text{So, the equation of the tangent line is} \\ y = -3(x+1)\ . \end{aligned}$$

Recall that “$+C$" means that we can add any constant to the function. Since $\tan^2 x = \sec^2 x - 1$, Students A and B have equivalent answers: they only differ by a constant.

So, if one is right, both are right; if one is wrong, both are wrong. We check Student A's work: $$\frac{\mathrm{d}}{\mathrm{d}x}\{\tan^2 x + x + C\}=\frac{\mathrm{d}}{\mathrm{d}x}\{\tan^2 x\} +1 + 0 = f(x)-1+1=f(x)$$ So, Student A's answer is indeed an anditerivative of $f(x)$. Therefore, both students ended up with the correct answer.

Remark: it is a frequent occurrence that equivalent answers might look quite different. As you are comparing your work to others', this is a good thing to keep in mind!

(a) When $x=3$,

$$\begin{aligned} F(3)&=\displaystyle\int_0^3 3^3 \sin(t)\ \mathrm{d}t=27\int_0^3 \sin t \ \mathrm{d}t \\ \text{Using the Fundamental Theorem of Calculus Part 2,} \\ &=27\left[-\cos t\right]_{t=0}^{t=3} =27\left[-\cos 3 - (-\cos 0)\right]\\ &=27(1-\cos 3 ) \end{aligned}$$ (b) Since the integration is with respect to $t$, the $x^3$ term can be moved outside the integral. That is: for the purposes of the integral, $x^3$ is a constant (although for the purposes of the derivative, it certainly is not).

$$\begin{aligned} F(x)&=\displaystyle\int_0^x x^3 \sin(t)\ \mathrm{d}t = x^3 \int_0^x \sin(t)\ \mathrm{d}t \\ \text{Using the product rule and the Fundamental Theorem of Calculus Part 1,} \\ F'(x)&=x^3\cdot \sin(x) + 3x^2 \int_0^x \sin(t)\ \mathrm{d}t\\ &=x^3\sin(x)+3x^2\left[-\cos(t)\right]_{t=0}^{t=x}\\ &=x^3\sin(x)+3x^2[-\cos(x)-(-\cos(0))]\\ &=x^3\sin (x) + 3x^2[1-\cos (x)] \end{aligned}$$

Remark: Since $x$ and $t$ play different roles in our problem, it's crucial that they have different names. This is one reason why we should avoid the common mistake of writing $\int_a^x f(x)\mathrm{d}x$ when we mean $\int_a^x f(t)\mathrm{d}t$.